\(\int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\) [907]

Optimal result

Integrand size = 23, antiderivative size = 117 \[ \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 b^2 d \sqrt {\sin ^2(c+d x)}} \] Output:

3*A*hypergeom([-1/6, 1/2],[5/6],cos(d*x+c)^2)*sin(d*x+c)/b/d/(b*cos(d*x+c) 
)^(1/3)/(sin(d*x+c)^2)^(1/2)-3/2*B*(b*cos(d*x+c))^(2/3)*hypergeom([1/3, 1/ 
2],[4/3],cos(d*x+c)^2)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.73 \[ \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=-\frac {3 \cot (c+d x) \left (-2 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right )+B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{2 d (b \cos (c+d x))^{4/3}} \] Input:

Integrate[(A + B*Cos[c + d*x])/(b*Cos[c + d*x])^(4/3),x]
 

Output:

(-3*Cot[c + d*x]*(-2*A*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2] + 
 B*Cos[c + d*x]*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2])*Sqrt[Sin 
[c + d*x]^2])/(2*d*(b*Cos[c + d*x])^(4/3))
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3227, 3042, 3122}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}}dx\)

\(\Big \downarrow \) 3227

\(\displaystyle A \int \frac {1}{(b \cos (c+d x))^{4/3}}dx+\frac {B \int \frac {1}{\sqrt [3]{b \cos (c+d x)}}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle A \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}}dx+\frac {B \int \frac {1}{\sqrt [3]{b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {3 A \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 b^2 d \sqrt {\sin ^2(c+d x)}}\)

Input:

Int[(A + B*Cos[c + d*x])/(b*Cos[c + d*x])^(4/3),x]
 

Output:

(3*A*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d* 
(b*Cos[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(2/3) 
*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(2*b^2*d*S 
qrt[Sin[c + d*x]^2])
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

Maple [F]

\[\int \frac {A +B \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}}}d x\]

Input:

int((A+B*cos(d*x+c))/(cos(d*x+c)*b)^(4/3),x)
 

Output:

int((A+B*cos(d*x+c))/(cos(d*x+c)*b)^(4/3),x)
 

Fricas [F]

\[ \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \] Input:

integrate((A+B*cos(d*x+c))/(b*cos(d*x+c))^(4/3),x, algorithm="fricas")
 

Output:

integral((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(2/3)/(b^2*cos(d*x + c)^2), 
 x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c))/(b*cos(d*x+c))**(4/3),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \] Input:

integrate((A+B*cos(d*x+c))/(b*cos(d*x+c))^(4/3),x, algorithm="maxima")
 

Output:

integrate((B*cos(d*x + c) + A)/(b*cos(d*x + c))^(4/3), x)
 

Giac [F]

\[ \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \] Input:

integrate((A+B*cos(d*x+c))/(b*cos(d*x+c))^(4/3),x, algorithm="giac")
 

Output:

integrate((B*cos(d*x + c) + A)/(b*cos(d*x + c))^(4/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}} \,d x \] Input:

int((A + B*cos(c + d*x))/(b*cos(c + d*x))^(4/3),x)
 

Output:

int((A + B*cos(c + d*x))/(b*cos(c + d*x))^(4/3), x)
 

Reduce [F]

\[ \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\frac {\left (\int \frac {1}{\cos \left (d x +c \right )^{\frac {1}{3}}}d x \right ) b +\left (\int \frac {1}{\cos \left (d x +c \right )^{\frac {4}{3}}}d x \right ) a}{b^{\frac {4}{3}}} \] Input:

int((A+B*cos(d*x+c))/(b*cos(d*x+c))^(4/3),x)
 

Output:

(int(1/cos(c + d*x)**(1/3),x)*b + int(1/(cos(c + d*x)**(1/3)*cos(c + d*x)) 
,x)*a)/(b**(1/3)*b)